QA 

59 


UC-NRLF 


B   M   EMT   2^D 


^%^ 


LOGARITHMS. 


^-^.^^ 


H.  N.  |WHEELER 


CAMBRIDGE : 

Charles  AV.  Sever,  University  Bookseller. 

1882. 


LOGARITHMS. 


BY 

H.  N.  WHEELER. 


3j®<0 


CAMBKIDGE: 
Charles  W.  Sever,  University  Bookseller. 

1882. 


LOGAEITHMS 


§  1,  The  logarithm  of  a  number  N  is  the  exponent  de- 
noting the  power  to  which  a  fixed  number  called  the  base 
must  be  raised  in  order  to  produce  N. 

Thus,  \i N=V,  then  is  log^N^x;  or,  in  words,  when 
the  base  is  h  the  loo;aritlim  of  N  is  x. 


What  is  logo 8? 

8  =  2^ ; 

.-.  log28  =  3. 

What  is  log.  — ? 

h^C^^ 

.•.log,f^  =  4. 

WhatisloggS? 

3  =  9^; 

•••log93  =  ^. 

What  is  log  10 1000? 

1000=103;   . 

•.  Iogiol000  =  3. 

What  is  log3  27? 

Ans.  3. 

What  is  log  4  64? 

Arts.  3. 

What  is  the  number  whose  loo-arithm  is  5  when  the  base 


is  2? 


log2-?v^=  5  ; 


N^'2^ 


What  is  the  number  whose  logarithm  is  3  when  the  base 
is  3?  Ans.  27. 

What  is  the  number  whose  logarithm  is  3  when  the  base 
is  10?  Ans.  1000. 

What  is  the  number  whose  logarithm  is  —  2  when  the 
baseisS?  log3^=-2;     .-.  iY=  8"^  =JL  =  JL. 


2  LOGARITHMS. 

What  is  the  number  whose  logarithm  is  —  2  when  the 
base  is  10?  ^.^^     1   , 

■  lOO' 
2 
What  is  the  number  whose  logarithm  is  —  -  when  the 

base  is  27?  ^^^g    1. 

■  9 

§  2.  The  logarithms  of  all  numbers  referred  to  the  same 
base  are  said  to  belong  to  the  same  system  of  logarithms. 
Thus,  log]o6,  log  10 7,  logio84,  logio768,  all  belong  to  the 
denar}-  or  common  system. 

§  3.  In  any  system  of  logarithms  the  logarithm  of  the 
base  itself  is  1 . 

For  2  =  2^     .-.logs  2    =1: 

10=101;   _._  iog,^iO=l: 

and,  in  general,  b  =  h^  \     .-.  logj?>     =  1 . 

§  4.  In  every  sj'stem  of  logarithms  the  logarithm  of  1  is  0. 
For  1=2";     .-.  loggl  =0: 

1  =  10";   .-.  logiol=0: 
and,  in  general,  \  —  lP\     .-.  log^l  =  0. 

§  5.  In  any  system  the  logarithm  of  the  reciprocal  of  a 
number  is  the  negative  of  the  logarithm  of  the  number. 

Proof:  —  is  the  reciprocal  of  If ;    now  -  =  h~'' ; 
If  If^ 

What  is  log.  —  ? 

l  =  i,  =  4-- 
IG      4- 

.*.  logi —  =  log4~-  =  —  2  :  or  log. —  =  —  log4lG  =  —2. 
°  16         °  16 


LOGAEITHMS. 


What  is  log  10 0.1?      0.1  =-i-  =  10-1;  ...  iog,o0.1  = -1. 


What  is  logs-;;? 


Ans. 


What  is  the  number  whose  logarithm  is  —  -  when  the 
base  is  16?  ^ 

16l      32 
What  is  the  number  whose  logarithm  is  —  3  when  the 
base  is  10?  Ans.  0.001. 


§  6.  In  any  s^'stem  the  logarithm  of  the  product  of  two 
or  more  numbers  is  equal  to  the  sum  of  the  logarithms  of 
the  numbers. 

Proof:    If  l  =  h%         m  =  6^         n  =  h% 

then  is  logjZ  =  .v,  log677r  =  i/,  \ogj,n  =  z', 

now  I  xm  xn  =  h''  X  b"  X  b^  =  /j^  +  ^  +  ^  ; 

.-.  log(?  X  m  X  n)  =  x-\-y  +  z  =  log?  +  logm  +  logn. 

It  is  evident  that  this  principle  ma}'  be  applied  to  any 
number  of  factors. 

What  is  log  2  (8  x4x  32)? 

8  =  2^  .-.  logo8    =3 
4  =  22;  _._  iog^4    ^2 
32  =  2^   .-.  log232  =  5 
logo(8x4x  32)  =  log28+log24+log232 

=  3  +  2+5  =  10. 
1 


Whatisloggf^  X  3  X  81  1? 


Ans.  -2  +  1+4  =  3. 


If  log  10 123  =  2.0899,  what  is  log ,o  12300? 

12300  =  100  X  123  =  10-  X  10-"«^  ; 

.-.  Iogaol2300  =  2 +  2.0899  =  4.0899. 

If  log  10 2  =  0.3010,  what  is  log lo  200?  Ans.  2.3010. 


4  LOGARITHMS. 

§  7.  In  any  system  the  logarithm  of  a  fraction  is  equal 
to  the  logarithm  of  the  numerator  mimis  the  logarithm  of 
the  denominator. 

Proof:   If  /  =  h''.  m  =  6^,  then  is  logj?  =  x,  log^m  =  y  ; 

lb''  I 

now  -  =  -  =  b^  " ;        .-.  \ogi-  =  X - y  =  logt,l -logtm. 
m      ¥  m 

What  is  logo  —  ? 

logo  — =  log,8-lo2:.32  =  3-5=  -2. 

Given:  logioS  =  0.-4771  and  log io2  =  0.3010  ;  what  is 
log  107?  Arts.  0.17G1. 

Given:  log^  123  =  2.0899  ;  what  is  logio0.123?_ 

Ans.  2.0899-3  =  1.0899. 

§  8.  In  any  system  the  logarithm  of  any  power  of  a 
number  is  equal  to  the  logarithm  of  the  number  multiplied 
by  the  exponent  of  the  poicer. 

Proof:   If  I  =  b\  then  is  log^Z  =  .r.  ?"■  =  (6^)"'  =  b"^ : 

.-.  logZ"  =  mx  —  m  X  logZ. 

Under  this  head  may  be  brought  a  root,  for  a  root  may 
be  regarded  as  a  fractional  power. 

7"  =  (6-)- =  5™;  .-.10??^  =  -=-  xlog?. 

^  ^  m      m         ° 

What  is  logo-i^?  log24«=  6  X  logoi  =  G  x  2  =  12. 

What  is  lo2,8*?  Iogo8*  =  ^log,8  =^  X  3  =  2. 

o-  3    °-        3 

WhatislogioA'TOO? 

A  Too  =  vTo^  =  lOf  :     . • .  log  10 <'Tob  =  |- 

What  is  log 2^/16"^?  Ans.  '^- 


LOGARITHMS, 


COMMON    LOGARITHMS. 

§  9.  The  logarithms  most  convenient  for  practical  use 
are  the  so-called  common  logarithms  in  which  the  base  is 
10.  The  following  table  contains  the  common  logarithms 
of  certain  integral  powers  of  10,  and  also  shows  between 
what  limits  the  logarithms  of  certain  other  numbers  must 
He. 


10000  =  10* ; 
9648  =  lO-'  +  a-iw™^; 
1000  =  l(f ; 

7g8  =  10-  ~  *  decimal . 

100  =  10- ; 

88  =   I0l^a.decmal. 

10  =  10^ ; 

J  zzz  l(jO  —  a  decimal . 

1  =  W ; 

0.66  =   10-1 -a  decimal. 

0.1=  — =10-^ 

lU 
0.079  =  10-2-^  a  decimal. 

0.01  =  — =  —=10- 

100         10^^ 
0.00684  =  10-3 -"decimal. 


0.001  = 


1 


=-L=io- 


1000     10-^ 

0.000795=  10-<H- a  decimal; 


0.0001  =- 


1 


10000     10* 


=-l.=io- 


log  10000  =  4. 

log  9648       =  .3  +  a  decimal. 


log  1000 

=  .3. 

log  768 

=  2  +  a  decimal. 

log  100 

=  2. 

log  88 

=  1  -^  a  decimal. 

log  10 

=  1. 

log  7 

=  0  +  a  decimal. 

logl 

=  0. 

log  0.66 

=  —  1  +  a  decimal. 

log  0.1 

=  -!• 

log  0.07 

=  —  2  +  a  decimaL 

log  0.01 

=  —  2. 

log  0.00684 

=  —  3  +  a  decimal. 

log  0.001 

=  -.3. 

log  0.000795=  —  4  -f  a  decimaL 
.log  0.0001= -4. 


6  LOGARITHMS. 

§  10.  From  §  9  we  see  that  the  logarithm  of  a  number 
which  is  not  tiu  integral  power  of  10  is  an  integer  plus  a 
decimal.  Take  9G48  for  example  ;  it  is  between  1000  [10'] 
and  10000  [10^],  therefoi'e  its  logarithm  is  between  3  and 
4,  i.e.  its  logarithm  is  3  +  a  decimal.  Take  0.000795  ;  it 
is  between  0.0001  [lO"'']  and  0.001  [10"=^],  therefore  its 
logarithm,  being  between  —  3  and  —  4,  can  be  written 
either  —  3  —  a  decimal,  or  —  4  +  a  decimal.  The  latter 
method  will  be  adopted. 

Suppose  the  decimal  part  of  the  logarithm  of  0.000795 
to  be  0.9004,  tlien  is  log 0.000795  =  -  4  +  0.9004.  Since 
we  shall  make  the  decimal  part  alwajs  +,  its  sign  need  not 
be  written ;  and  in  order  to  denote  that  the  negative  sign 
applies  to  the  integral  part  onl}',  we  write  it  above  the 
integer  ;  thus,  log  0.000795  =  4.9004. 

Let  us  take  an  exami)le  to  show  the  advantage  to  be 
gained  by  making  the  decimal  part  of  the  logarithm  always 
positive. 

Suppose  log72G  =  2.8009  ;   find  logO.0726. 

0.072G  =  -I^  =  I^; 
10000       10* 

.-.logo. 0726  =  log  726 -log  10*  =  2. 8009-4  =  2.8609. 

Comparing  tliis  result  with  the  given  logarithm,  we  see 
that  the  decimal  parts  are  exactly  the  same  ;  this  will  not 
be  the  case  if  log  0.0726  is  expressed  in  the  other  way,  for 
tlien  we  shall  get 

-2  +  0. 8609  =  -1.1391  or  -1-0.1391. 

Let  us  take  another  example. 
Given :   log  0.0726  =  2.8609  ;     find  log  7.26. 
7.26  =  6.0726  X  100; 

.-.  log  7.26  =  2.8609  +  2  =  0.8609  ; 


LOGARITHMS.  7 

and  figain  we  see  that  the  decimal  part  of  the  result  is  the 
same  as  that  of  the  given  logarithm.  If,  then,  we  alwa3-s 
make  the  decimal  part  of  a  logarithm  +,  the  following 
statement  will  be  true  :  — 

Numbers  icliich  consist  of  the  same  se^'ies  of  figures,  i.e., 
which  differ  only  in  resjject  to  the  position  of  the  decimal 
2yoint,  ivill  have  the  same  decimal  ^xirts  to  their  logarithms ; 
and  the  integral  parts  of  the  logarithms  ivill  depend  upon 
the  position  of  the  decimal  point;  for  a  change  in  the  posi- 
tion of  the  decimal  point  is  equivalent  merely-  to  multiply- 
ing or  dividing  by  an  integral  power  of  10,  and  this  process 
will  only  increase  or  diminish  the  logarithm  (exponent)  by 
an  integer. 

§  11.  The  integral  part  of  a  logarithm  is  called  its 
characteristic,  and  the  decimal  part  its  mantissa. 

The  characteristic,  when  the  number  is  not  an  integral 
power  of  10,  Ynmy  be  found  easil}^  by  noting  between  what 
adjacent  integral  powers  of  10  the  number  lies. 

Thus,  9648  (see  §  9)  lies  between  10^  [1000]  and  10* 
[10000]  ;  therefore  the  characteristic  of  log9G48  is  3  : 

0.0007G5  lies  between  0.0001  [10-*]  and  0.001  [10-^]  ; 
therefore  the  characteristic  of  log  0.000765  is  —4. 

Thus  we  see  that  the  characteristic  of  the  logarithm  of  a 
number  depends  wholly  on  the  })osition  of  the  decimal 
point,  and  not  at  all  on  the  figures  which  the  number 
contains. 

From  what  precedes  we  can  readily  deduce  the  following 
rule  for  finding  the  characteristic  of  the  logarithm  of  any 
number :  — 

The  characteristic  of  the  logarithm  of  a  number  is  equal 
to  the  number  of  places  by  which  the  first  significant  figure 
of  the  number  is  removed  from  the  units'  place,,  and  is  posir 


8  LOGARITHMS. 

live  for  numbers  greater  than  unity  (i.e.,  where  the  first 
significant  figure  is  to  the  left  of  the  decimal  2'>oint) ,  and 
negative  for  mimhers  less  than  iinity  (i.e.,  tchere  the  first 
significant  figure  is  to  the  right  of  the  decimal  2'>oint)  ;  it  is 
0  whe7i  the  first  significant  figure  is  in  the  units'  place. 

Examjyles:  Find  by  the  above  rule  the  characteristics 
of  the  logarithms  of  the  following  numbers  :  — 

3  2  10  2  10  0 

(1)1689;  (2)168.9;  (3)1.689; 

(4)0.'l689;       (o)0"o~i689;      (6)  o7o~o'o'l689. 

Answers.    (1)3;        (2)  2;       (3)0; 
(4)-l;   (5)-2;   (6)-4. 

§  12.  The  mantissas.,  or  decimal  parts  of  the  logarithms 
of  numbers,  have  been  computed  to  different  degrees  of 
accurac}',  the  results  for  some  numbers  having  been  carried 
out  to  more  than  25  places  of  decimals.  A  collection  of 
these  mantissas  is  called  a  logarithm-tahle.  Tables  are 
characterized  by  putting  before  the  word  logarithm  the 
expression  Z-place,  4-place,  5-place,  1-place.,  d-jylace,  etc., 
according  to  the  number  of  places  of  decimals  to  which  the 
results  are  given.  Thus,  a  4-place  logarithm-table  is  one 
in  which  the  results  are  given  to  four  places  of  decimals. 

§  13.   Examples  of  finding  logarithms  from  the  tables. 

Find  log  495000. 

Referring  to  p.  2  of  Peirce's  Tables,  we  seek  49  at  the 
left  of  the  table  in  the  column  headed  n,  and  run  along 
the  horizontal  line  in  which  this  lies  until  we  reach  the  ver- 
tical column  which  has  5  at  the  top,  where  we  find  6946, 
and  this  is  the  mantissa  or  decimal  part  (accurate  to  4 


LOGARITHMS. 


9 


places)  of  the  logarithms  of  all  numbers  whose  significant 
figures  form  the  series  495  ; 

.-.  mantissa  of  log  495000  =  .G946. 
By  the  rule  of  §  11  the  characteristic  is  5  ; 

-      .-.  log 495000  =  5.6946. 
Find  the  logarithms  of 
(1)  9.84;  (2)  0.729;  (3)  6690; 

(4)  0.066  ;  (5)  0.00000543  ;  (6)  0.007. 

Answers.    (1)0.9930;    (2)1.8627;   (3)3.8254; 
(4)  2.8195;    (5)  6.7348;   (6)  3.8451. 

What  is  log  241 50? 

"We  do  not  find  this  number  in  the  table  ;  we  do  find, 
however,  log  24100  =  4.3820, 

and  log  24200  =  4.3838. 

The  given  number  24150  is  half-way  between  24100  and 
24200  ;  i.e.,  it  is  equal  to  the  first  of  these  numbers  plus 
half  the  distance  between  the  first  and  second.  Assuming* 
that  the  required  logarithm  is  half-wa}'  between  4.3820  and 

*  The  assumption  that  numbers  vary  proportionally  with  their 
logarithms  — that  in  the  above  example,  for  instance,  if  a  num- 
ber is  half-way  between  two  numbers,  its  logarithm  is  half-way 
between  the  logarithms  of  these  numbers  —  is  only  approxi- 
mately correct ;  but,  when  the  difference  between  the  numbers 
is  small,  the  error  to  which  this  assumption  leads  us  is  smaU; 
and  in  every  properly-constructed  table  the  differences  or  inter- 
vals between  the  successive  numbers  of  the  table  are  so  small 
that  the  error  in  a  logarithm  fouud  as  above  will  not  affect  the 
last  place  of  decimals. 

This  method  of  finding  the  logarithm  of  a  number  which  lies 
between  two  successive  numbers  of  a  table  is  called  the  method 
of  interpolation. 

If  we  refer  to  a  7-place  table,  we  find  from  the  table,  without 
interpolation,  log  24150  =  4.3829171,  and  this  carried  to  only  4 
places  of  decimals  is  4.3829,  which  is  the  result  obtained  from 
the  4-place  table  by  interpolation. 


to 

4.383&,  i.e.  Hiatiki&e^paLto  thfrSrsfr  of  these  losarrthms 
ploji  half  the  cS&can£&  fisam.  th&  first  1og,irtt.hin  to  thfr  see-^ 
Offli,  we  gpc 

We  csfflc  if  we  wish*.  S»i  .5  of  O.OOl!^  tbont  the  pattr  i  )i" 
tijtj  loi^attthm  oi"  itlOO  uWiLwe  riauih.  tha  ^t-r  tun 


'Wba^  is-  lo^:5t>7T 

T^  5680 


*ttct  the  dttfertjiKti  ,s  '-^  5.  'tjor. 


V.,,.  ,. 


12  LOGARITHMS. 

Find  the  logarithms  of 

(1)10.09;     (2)190.1G;     (3)1000.4;      (4)0.12168. 
Answers.   (1)  1.0039  ;   (2)  2.2791  ; 
(3)  3.0002  ;  (4)  1.0852. 


Examples  : 

Find  the  logai 

ithms  of  the  following  numbers  : 

1.834 

19.98 

0.01592 

1.2899 

54.97 

1587.1 

10.041 

1000 

9990 

9999 

0.09998 

0.00000000010007 

79930000     0.0001001 

50090 

0.6394. 

Ansivers  : 

0.2G34 

1.3006 

2.2019 

0.1106 

1.7402 

3.2006 

1.0018 

3.0000 

3.999G 

4.0000 

2.9999 

l0.0003_ 

7.9027 

4.0004 

4.6997 

f.8058.  ' 

§  14.  To  find  the  number  ichich  corresjyonds  to  a  given 
logarithm. 

Of  what  number  is  1.8156  the  logarithm? 

We  find  the  mantissa  .8156  on  p.  3,  in  the  same  hori- 
zontal line  witli  65  (on  the  left)  and  in  the  vertical  column 
headed  4,  therefore  the  significant  figures  of  our  number 
are  654  ;  now  the  characteristic  of  our  logarithm  being  1, 
the  first  significant  figure  6  of  our  number  must  be  the  first 
to  the  left  of  the  units'  place,  hence  5  is  in  the  units'  place, 
and  the  required  number  is  65.4. 

Find  the  munbers  whose  logarithms  are 

(1)  2.7789;    (2)  4.3160;    (3)  0.5478. 

Ansivers.    (1)  0.0601;   (2)  20700;   (3)  3.530. 

Find  the  number  (iV)  Avhose  logaritlim  is  0.5506. 

We  do  not  find  tliis  logarithm  in  the  table  ;  we  do  find, 
however,  two  successive  logarithms  between  which  this 
lies.     Thus, 


LOGARITHMS.  13 

log  3.550  =  0.5502, 
logN       =0.5506, 
log  3.560  =  0.5514. 
The  difference  between  0.5502  and  0.5506  is  4  (in  the  4th 
place),  and  the  difference  between  0.5502  and  0.5514  is  12  ; 
.-.  logiV^=log3.550+y42.of  (log 3.560 -log 3.550); 
hence,  by  the  method  of  interpolation, 

iVr=  3.550  +  y%  (3.560  -  3.550) 

=  3.550  +  -j^  (0.010)  =  3.550  +  0.003^ 
=  3.553= 
^  of  10  may  be  found  in  the  table  of  proportional  parts  ; 
thus,  in  the  same  horizontal  line  with  our  logarithm  we 
find  4,  and  vertically  above  this  at  the  top  we  find  3,  which 
is  the  fourth  significant  figure  of  the  number  whose  log- 
arithm exceeds  5502  by  4. 

Given:  log.V=  3.5449  ;   find  N. 

Seeking  in  the  table  the  logarithm  Clearest  the  given 
logarithm,  we  find  log  0.003510  =  3.5453,  which  exceeds 
the  given  logarithm  l;)y  4  ;  and  aliove  4  in  the  table  of  pro- 
portional parts  we  find  3,  which  is  in  tliis  case  to  be  sub- 
tracted from  the  fourth  significant  figure  of  the  number 
whose  logarithm  is  3.5453  ; 

.-.  iVr=  0.003510  — 0.000003  =  0.003507. 

Examples :  Find  the  numbers  which  correspond  to  the 
following  logarithms :  — 


1.8359 

2.4089 

1.8058 

3.4429 

3.4631 

4.0003 

4. 099 7 

9.9016 

1.0020 

1.1109 

0.0001 

3.7402. 

Ansivers  : 

68.53 

0.025G4 

0.6394 

2773 

0.002905 

0.00010008 

50090 

0.000000007972 

10.046 

12.91 

1.0002 

5498. 

14 


LOGARITHMS. 


§  15.     Examples  of  Products  ill ast rating  §  G. 
1.    Find  the  product  of  78.05,  0.6178,  341U0,  10.009, 
and  0.0009. 

Solution :  We  find  by  the  tables, 


log  78.05 
logO.G178 
log 34 100 
log  10.009 
los  0.0009 


1.8924 

9.7909  —  10* 
4.5328 
1.0004 
0.1)542  -  10 


log  of  prod.  =  24.1707  -  20  =  4.1707  ; 

.-.  the  requh'ed  product  is  14815. 

*In  order  to  avoid  the  use  of  negative  characteristics,  com- 
puters generally  add  10  to  each  negative  characteristic,  and  then 
write  —10  after  the  logarithm. 

2.    Find  the  product  of 

(1)  4.31,  0.39,  G4,  and  1.02; 

(2)  3G1,  0.043,  and  0.00621  ; 

(3)  3.81,  97.6,  3120,  and  0.00081  ; 

(4)  2.358,  4.321,  0.8765,  and  1.12. 

Aiisicers.    (1)  109.75;   (2)  0.09G4  ; 
(3)  939.8;      (4)  10. 

§  16.  The  Arithmetical  Complement  of  a  logarithm^ 
and  examples  of  reciprocals  illxstrating  §  5. 

The  arithmetical  complement  of  a  logarithm  —  written 
colog  —  is  tlio  remainder  which  we  get  after  subtracting  the 
logarilhui  IVoui  10. 

Thus,  if  log78.05  =  1.8924, 

then  is  colog  78.05  =  10-1 .8924 

=  8.1076. 
Find  colog  6.912. 

log6.912  =  0.8396; 

.-.  colo<r6.912  =  9.1604. 


LOGARITHMS.  15 

The  arithmetical  complement  of  a  given  logarithm  may 
be  obtained  with  great  ease  if  we  note  that  subtracting  a 
logarithm  from  10  is  equivalent  to  subtracting  the  last  sig- 
nificant figure  from  10,  and  each  of  the  other  figures  from 
9.     In  getting  colog6.912,  for  instance, 

10.0000 
loa:6.912  =  0.8396 


colog6.912  =  9.1004 
we  can  begin  on  the  left,  and  saj,  0  from  9  is  9,  8  from  9 
is  1,  3  from  9  is  6,  9  from  9  is  0,  and  6  from  10  is  4. 

Findcologsof  (1)  7.3G  ;  (2)  0.0134  ;  (3)  81.2  ;  (4)  3041. 
Ansivers.    (1)  9.1331;   (2)  11.8729; 
(3)  8.0904;   (4)  6.5170. 
We  have,  in  general, 

colog^=  10  —  logiV; 
subtracting  10  from  both  sides  of  this  equation  we  get 

colog^—  10  =  —  log^, 
and  from  tliis  we  see  that  cokxj  N  diminislied  by  10  is  the 
same  as  the  negative  of  log  N. 
From  §  5  we  have 

log  —  =  —  log^=  (by  the  above)  cologiV—  10. 

Hence,  the  logarithm  of  the  reciprocal  of  a  number  is  equal 
to  the  arithmetical  complement  of  the  logarithm  diminished 
by  10. 

Find  losjc — •• 
"273 

log—  =  colog273  -  10  =  7.5638  -  10  =  3.5638. 
*273 

Using  the  colog  here  has  enabled  us  to  obtain  very  easily 

a  positive  mantissa  for  loa; 

^  ^273 


16  LOGARITHMS. 


Find  the  values  of 

(1)^'  (2)  Jt^;  (3):^;    (4) 


319'    ^  ^  0.0173'    ^   '  36.18'     ^  '   0.U00123 

Answers.    (1)  0.003135;    (2)  57.81  ; 
(3)  0.02764;    (4)  8130. 

§  17.     Examples  o/ Fractions  illustrating  §  7. 
1.    Divide  0.01478  by  0.9243. 
Solution:    We  have,  by  the  tables, 

log0.01478  =  8.1697 -10 
logo. 9243    =9.9658-10 


subtracting,  we  get  log  of  quotient  =  2.2039 

.-.  the  required  quotient  is  0.01599. 

Remembering  that  cologiV"— 10  = —log ^^,   we  can   do 
the  above  example  as  follows  : 

logo. 01478  =8.1697-10 

colos  0.9243  -  10  =  0.0342 


and  adding,  we  get  log  of  quotient  =  8.2039  —  10 

=  2.2039 

„     T-    1  ^-1         1         P         894x0.17.S6x41 
2.    Innd  the  value  of 


0.86x0.0007952x9001 


log894        =    2.9513  logO.86  =    9.9345-10 

logo. 1786=    9.2519-10     logO.0007952  =    6.9005-10 
log41         =    1.6128  los9001  =    3.9542 


log  numer.  =  13.8160  -  10     logdeuom.       =  20.7892  -  20 
=    3.8160  =    0.7892 

0.7892 


.logquot.  =    3.0268     and  tlie  required  quotient  is  1063.8. 


LOGARITHMS.  17 

We  will  now  do  the  above  by  the  aid  of  cologs. 

log894  =  2.9513 

logo. 1786  =  9.2519-10 

log41  =  1.6128 

cologO.86-10  =  0.0655 

colog 0.0007952  -  10  =  3.0995 

»      coloo-9001  -  10  =  6.0458  -  10 


23.0268  -  20 
=    3.0268 
A  comparison  of  these  two  methods  of  solution  shows 
the  advantage  to  be  gained  by  the  use  of  cologs. 

Find  the  values  of 

.J.    3176  X  4.31  X  .023.        .,-,.    731x9.16. 
^   ^  6.4 X. 231  '        ^"^    2113x27  ' 

.g.    6. 19  X  37000  X. 002  .      ,^.    49880  X  .03754  X  68.1  . 
^  ^         31.96x40.1        '     ^  ^    7.816x578.9x28.43' 
A7iswers.    (1)212.9;      (2)0.1174; 
(3)  0.3575  ;    (4)  0.9910. 

§  18.     Examples  o/ Powers  and  Roots  illustrating  §  8. 

1.  Find  the  fourth  power  of  0.9857. 
Solution :   We  have  b}'  the  tables, 

log   0.9857     =    9.9937-10 
multiply  by  4 

log  (0.9857)*  =  39.9748  -  40  =  1.9748  ; 

.-.  (0.9857)^  =  0.9436. 

2.  Find  the  values  of 

(1)  (7.013)^  (2)  (.3009)^ 

(3)  (1.04)1^  (4)  (2.314)«. 

Answers.    (1)344.9;    (2)  .002469 ; 
(3)  1.663;    (4)  153.4. 


18  LOGARITHMS. 

3.  Find  the  cube  root  of  G3. 
Solution :   We  have  bj-  the  tables, 

log  63  =1.7903, 
dividing  this  b^-  3  (multiplying  l)y  i)  we  get 

logv'63  =  0.5998; 

.-.  ^v/63=:  3.979. 

4.  Find  the  fifth  root  of  0.02814. 

Solution :   We  have  by  the  tables, 
logO.02814  =  2.4493  ; 

if  we  divide  the  logarithm  by  5  we  get  —^  for  a  character- 
istic ;  we  will  therefore  increase  tlie  characteristic  —  2  by 
50*  and  write  —50  after  tlie  logarithm  ;  thus, 

logo. 02814  =  48.4493  -  50  ; 
now,  di\ading  by  5,  we  get 


log  VO. 02814  =  9.6899  -  10 
=  1.6899; 


.-.  V0.02814  =  0.4897. 

*  A  number  of  tens  equal  to  the  exponent  of  the  root ;  i.e.,  in 
this  example,  5  teiK,  or  50. 

5.  Find  the  cube  root  of  0.27. 

Iog0.27  =  1.4314; 
increasing  the  characteristic  —1  b}'  3  tens,  or  30,  we  get 
logo. 27      =29.4314-30; 
.-.log  ^0.27=    9.8105-10 
=    1.8105; 

.•.-v/o:27  =  0.6464. 

6.  Find  log  (0.1 684)1. 

First  find  log  (0.1684)-,  and  then  divide  this  logarithm 
by  3.  Ans.   0.3049. 


LOGARITHMS. 


19 


Find  the  values  of 


7.1;         (2)V:0041;       (3)V3l430; 
{A)A/lAm;     (5)^v^37J:9;       (6) -^0.0000492. 
Answers.    (1)  1.922;   (2)  0.2530;   (3)  7.933; 
(4)  1.072;   (5)  1.321;     (G)  0.3321. 


§  19.     Miscellaneous  Problems  and  Examples. 

1 .  Given  the  exponential  equation  20  =  2'' ;  find  x. 
By  §  8  we  have 

log  20  =  a;  X  log  2  ; 
,^log20^  1.3010 
*  '  ^~  log 2  ~  0.3010' 

.  • .  log  X    =  log  (log  20 )  —  log  (log  2 ) 
=  log  1.3010 -logo. 3010. 
logl.3010  =  0.1143 
logo. 3010  =  1.4786 

logo;  =0.6357 

.-.  a;  =  4.322. 

We  see  from  the  definition  of  a  logarithm  (§  1)  that  the 
above  value  of  x  is  the  logarithm  of  20  to  the  base  2  ; 

.-.  Iog2  20  =  4.322. 

We  can  in  like  manner  work  out,  by  the  aid  of  common 
logarithms,  the  logarithms  of  all  numbers  to  any  base. 

2.  Find  loggO.G. 

Put  0.6  =  2^;   .-.  is  logO.G  =a;  X  log2; 

.^  logo. 6^1.7782 
'  '  '^~   iog2    ~  0.3010' 

In  order  to  obtain  the  value  of  x  from  this  equation,  we 
must  observe  that  the  form  1.7782,  although  sometimes 
convenient  for  purposes  of  addition  and  subtraction,  is  not 


20  LOGARITHMS. 

suitable  for  purposes  of  inultipliea,tion  or  division  ;  we  will 
therefo]"c  reduce  it  to  its  equivalent, 

-1  +0.7782=  -0.2218; 

^0.2218' 


._  -0.2218 
^~    0.3010 


o.;30io 


The  value  of  the  part  in  brackets,  being  positive,  can  be 
found  in  the  usual  manner. 

Thus,  logo. 2218  =  1.3400 

los0.3010  =  T.478G 


1.SG74 

0  2-?18 
..-.  ^-^        =0.7368; 
0.3010 

.-.  x=  —  0.73G8. 

*  Negative  iiumljcrs  have  uo  real  logarithms  ;  for  we  can  give 
to  10  no  exponent  whicli  will  produce  a  negative  number,  since 
10  with  any  exponent  (positive  or  negative)  gives  always  a  pos- 
itive quantity,  greater  than  uuitj'  if  the  exponent  is  positive,  and 
between  zero  and  unity  if  the  exponent  is  negative. 

3.  The  number  2.7183  is  the  base  of  the  so-called  natu- 
ral system  of  logarithms,  a  sj'stem  much  used  in  the  higher 
mathematics. 

Find,  Avith  this  number  for  a  base,  the  logarithms  of 
0.7,  0.8,  and  0.9. 

Answers.    —  0.35G6,  —0.2231,  and  —0.1055. 

4.  How  many  digits  are  there  in  the  integral  part  of 

(1.04)«»°? 

Put  (1.04  )*■'"«'  =.V, 

then  6000  log  1 .  04    =  log  X, 

GOOO  X  0.0170  =  102.00=  log.V. 

The  first  significant  figure  of  X  is  removed  102  steps 
from  the  imits'  place,  and  there  are  thei'eforc  103  digits  in 
the  integral  part  of  JSf. 


LOGARITHMS. 


21 


5.  Show  that  there  is  onl}'  one  digit  in  the  integral  part 
of  (1.02y\ 

6.  Show  that  there  are  20  digits  in  2^\ 

7.  If  $1  be  put  at  d(/o  interest,  in  how  many  years  will 
it  amount  to  $2,  if  the  interest  is  compounded  annually? 

The  amount  at  the  end  of  one  year  is  $1.03  ;  this  amoimt 
now  forms  a  new  principal,  so  tliat,  at  the  end  of  the  sec- 
ond year,  ever>/  dollar  of  this  $1.0;l  amounts  in  turn  to 
$1.08,  and  the  tvJtoIe  amount  is  $(1.03)  (1.03)  =  $(1.03)^. 

At  the  end  of  the  third  3'ear  the  amotuit  is  evidently 
$(1.03)%  and  at  the  end  of  m  years,  $(1.03)'". 

In  this  example  we  are  required  to  find  a  value  m  such 
that 


(1.03)- 
mlosrl.OS 


2, 
log  2; 
Tog  2 


0.3010 


=  23.5. 


log  1.03      0.0128 
In  general,  if  $P  be  put  at  interest  at  a  I'ate  r,  com- 
pounded annuall}',  the  amount  at  the  end  of  m  years  will 

^®  $^=:$P(l+r)"'. 

8.  In  how  many  years  Avill  $1000  amount  to  $1800,  at 
5  per  cent  compound  interest? 

1800  =  1000(1.05)"'; 

log  1-8        10  1 

•'•  "^^^^i      1  n'  =  1^'  "early. 
logl.Oo 

9.  Show  that  a  sura  of  mone}'  at  compound  interest  will 
double  itself  in  al)out  7:}-  years  if  the  rate  be  10  per  cent, 
and  in  about  1  7t}  years  if  the  rate  be  4  per  cent. 

10.  Show  that,  if  the  population  of  a  given  State  increases 
by  o'oth  of  itself  in  every  year,  tlie  number  of  inhabitants 
at  the  end  of  14.2  years  will  l)e  twice  as  great  as  it  was  at 
the  beofinnins:  of  the  interval. 


■"«'^*'*4<*niiv«wv*ivMii>iimi 


R'Wf.WPflKD.'iaiJ/^^P';*' 


22  LOGARITHiVIS. 

[11.*  logO. 

—  =  10-^;  .-.   loii  —=-X. 

Let  a;  =  CO  (ineauiug  by  this  that  x  is  a  vai"ial)le  quantity  which 
is  supposed  to  iucrease  iucleflnitely,  tluit  is,  to  increase  without 
limit),  theu  lO-'  will  become  oo,  and  we  shall  have  loi;;—  =  —  oo. 

— ,  or  —  when  x  =  oo,  is  not  0 ;   it  is  a  quantity  which  may  be 

made  to  approach  0  as  nearly  as  we  please,  but  it  can  never  be 
made  to  reach  0;  such  a  quantity  is  called  an  infinitesimal,  and 
is  denoted  by  some  authors  by  the  symbol  o.  (I  believe  that 
this  notation  was  first  introduced  by  Professor  C.  H.  Judson  of 
South  Carolina.) 

Zero  has  no  logarithm;  many  authors  say,  however,  that  the 
logarithm  of  zero  is  —  co,  meaning  tliereb}^,  perhaps,  that  —  oo  is 
the  logarithm  of  an  infinitesimal,  that  is,  of  a  variable  which  has 
0  for  its  limit. 

I  very  much  prefer  to  use  some  symbol — a  liorlzontal  zero, 
for  instance  —  for  an  infinitesimal;  I  can  then  say  with  strict 
accuracy  that  log  o  =  —  co.  I  also  express  this  by  ])utting 
logO  =  — ob,  using  the  dot  over  the  sign  of  infinity  to  denote 
that  log  0  is  impossible,  and  retaining  the  symbol  —  »  because 
it  expresses  the  logarithm  of  a  quantity  whose  limit  is  0.  I  be- 
lieve that  this  mode  of  expression  can  be  used  advantageously 
in  other  cases,  and  that  its  adoption,  or  the  adoption  of  some 
other  method  whereby  we  are  not  compelled  to  regard  a  vari- 
able as  reaching  its  limit,  will  make  pcrfectlj^  clear  to  the  stu- 
dent of  mathematics  many  things  which  he  now  accepts  on 
faith. 

The  following  are  examples  of  cases  to  which  I  would  apply 
what  has  been  stated  above  :  — 

(1)  i  =  00  [not   i  =   ccl  ;     i  =  ob. 

(2)  00  X  o  =  rt,  not  00  X  0  —  «,  for  od  x  0  =  0. 

(3)  tan  90°  =  ob,  not  tan  90°  =  oo. 

(4)  Two  parallel  lines  do  not  meet  at  infinity ;  thev  may  be, 
however,  the  limits  approached  by  two  lines  whose  point  of  inter- 
section recedes  indefinitely.] 

*  This  is  intemled  lor  llie  toaclu'r  r;itlK'r  lliaii  lor  tlie  stiuioiit. 


LOGARITHMS. 


9?. 


12.    Find  the  value  of 

a/0.0432  X  (3.122)f  X  (0.0001)^' 
(18G2)2x  syo,0048 


logo. 0432       =58.6355-60    logVO.0432  =   9.7726-10 

[log3.122         =   0.4945 

ilog(3.122)«    =   3.9560  log(3. 122)7  =   0.5651 

logo. 0001  =   4.0000  log(0. 0001)1=    1.0000 

nogl862  =   3.2700 

!log(1862)-  =   6.5400  colog(1862)--10=   3.4600-10 

I  log 0.0048  =   3.6812 

(log  ^0.0048  =  1.2271  cologSyo.0048  - 10=  10.7729  - 10 

23.5706—30 
or       7.5706 
.-.  the  reqiured  value  is  0.0000003721. 

Find  the  values  of 

13.   ({O-^O'T^m  Ans.  0.5549. 


V0.4x67 

14.     W       (8.763)-^  XIQO;^^  ^^^^_  g_ 

>'9X  VO.1109  X  (4.9)i 


15.    ^/(0.01-2)-x0.27.  ^4,^^._  ,_o,3i_ 

\(64)2x  0.00651 


16.    Jv-27.4  +  (0.9)^^  ^^^^    jg4_97_ 

^^       0.0001021 

17     1^(134.9)- X  a/TgY  Ans.  0.0003164. 

V  10000  X  46.49  / 


18. 


0.001019004_x^0.j999  ^^^_  0.000004442. 

760  X  a/0.02751 


24  LOGARITHMS. 

§  20.     The  Trigonometric  Functions. 

The  trigonometric  functions  of  angles  between  0°  and  90°, 
at  intervals  of  10',  are  given  in  the  tables  on  pages  22-27. 
Let  us  find  the  functions  of  15°  12'  for  example.  Under 
<f>,  in  the  left-hand  column  of  the  tJmxl  division  of  page  22, 
we  find  15°  immediately  followed  b}-  10' ;  in  the  same 
horizontal  line  with  the  latter,  and  in  the  column  headed 
sin  </),  we  find  .2616,  which  is  sin  15°  10'. 

.-.  sin  15°  10'  =  0.2616,     and  similarly 
sin  15°  20' =0.2644 


difference  =        28. 
Now,  by  the  method  of  interpolation  (see  §  13),  the 
amount    to   be   added    to   0.2616    in    order    to    produce 
sin  15°  12'  is  .2  of  28,  or  5.6,  which  we  call. 6; 

.  .  sm  lo    12'  =  -^       ,,,,    , 
-f  .0006 


=      0.2622. 
In  like  manner,  from  page  22  and  from  pages  24  and  26, 
cos  15°  12'=  0.9650* 

tanl5°  12' =  0.2717  sec  15°  12' =  1.036 

ctn  15°  12'  =  3.681  esc  15°  12'  =  3.814. 

By  the  above  method  we  find  :  — 

sin  13°  28'  =  0.2328  sin  28°  13'  =  0.4728 

cos  13°  28'  =  0.9725  cos  28°  13'  =  0.8812 

*  Cos  15°  20'  being  less  than  cos  15°  10',  .2  of  the  difference 
was  subtracted  from  cos  15°  10'.  The  student  should  carefully 
consider  whether  the  amount  obtained  by  the  method  of  inter- 
polation is  to  be  added  or  subtracted. 

We  learu,  either  from  Trigonometry  or  from  an  examination 
of  the  tables,  that  the  sine,  tangent,  aud  secant  iiicrease,  and 
that  the  cosine,  cotangent,  and  cosecant  decrease,  as  the  angle 
increases  from  0°  to  90°. 


LOGARITHMS.  25 

tan  13°  28'  =  0.2395  tan  28°  13'  =  0.5365 

ctn  13°  28' =  4.176  ctn  28°  13' =  1.864 

sec  13°  28' =  1.028  see  28°  13' =  1.135 

CSC  13°  28'  =  4.294  esc  28°  13'  =  2.115, 

and  in  a  similar  way  are  to  be  found  the  functions  of  all 
angles  heticeen  0°  ayid  45°. 

§  21.  We  will  now  show  how  to  find  the  trigonometric 
functions  of  angles  l)etween  45°  and  90°. 

Let  us  take  58°  24'  for  example.  At  the  bottom  of  the 
right-hand  column  of  the  middle  division  of  page  23  we 
find  0  ;  we  run  up  this  column  luitil  we  find  58°,  and  im- 
mediately above  this  10',  20',  etc.  ;  in  the  same  horizontal 
line  with  20',  and  in  the  column  which  has  sin^  at  the  hot- 
torn^  we  find  .8511,  which  is  the  sine  of  58°  20',  Now,  hy 
the  method  of  interpolation,  we  get  sin  58°  24'  =  0.8517. 

Similarl}^ :  — 

cos  58°  24'  =  0.524  sec  58°  24'  =  1.909 

tan  58°  24'  =  1.625  ctn  58°  24'  =  0.6152 

CSC  58°  24'=  1.174 

sin  84°  13'  =  0.9949  sin  46°  28'  =  0.7250 

cos  84°  13'  =  0.10077  cos  46°  2S'  =  0.6888 

tan 84°  13'  =  9.88  tan  46°  28'  =  1.053 

ctn  84°  13'  =  0.10128  ctn  46°  28'  =  0.9501 

sec  84°  13'  =  9.93  sec  46°  28'  =  1.452 

CSC  84°  13'  =  1.0051  CSC  46°  28'  =  1.380. 

From  the  above  we  see  that  when  the  angle  is  between 
45°  and  90°  it  is  to  be  sought  in  the  right-hand  column, 
that  the  name  of  the  required  function  is  to  be  found  at 
the  bottom  of  the  page,  and  that  the  columns  are  to  be 
read  upwards. 


26  LOGARITHMS. 

§  22.  When  a  function  of  an  angle  is  given,  the  angle 
can  be  fonnd  from  the  tables  l^y  a  method  simihir  to  that 
alread}'  explained  for  finding  a  nnmber  from  its  logarithm. 
Thus,  if  sinvl  =  0.G792,  we  find  on  page  23, 

sin  42°  40'=0.G777 

sin  42°  50' =0.6799 


difference  =  0.0022 

and  sin  42°  50'-  sin^l  =  0.0007  ; 

.-.  ^  =  42°  50'-— X  10'  =  42°47'. 
22 

Examples :  Find  A  from  the  tables  in  each  of  the  fol- 
lowing cases : — 

(1)  sin  ^=0.9621;  (4)   sec  ^1  =  1.111  ; 

(2)  cos^l  =  0.12G8;  (5)   ctnyl  =  0.0079  ; 

(3)  tanJ.  =  2.172;  (6)   esc  J.  =  9.872. 
Answers:   (1)  74°  10';    (3)  65°  17';    (5)  89°  27' ; 

(2)  82°  43';    (4)  25°  50' ;    (G)     5°  49'. 

§  23.  Referring  to  the  tables,  pages  22-27,  we  notice 
that  the  angles  in  the  right-hand  columns  are  the  comple- 
ments of  the  angles  in  the  left-hand  columns.  Thus,  oppo- 
site 41°  10'  on  the  left  we  find  48°  50'  on  the  right,  and  we 

also  find 

sin  41°  10'  =  0.G583  =  cos  48°  50' 

cos 41°  10'  =  0.7528  =  sin  48°  50'. 

These  results  are  in  accordance  with  the  principles  of 

Trigonometry,  for  Trigonometry  tells  us  that  the  functions 

of  any  angle  (<^)  are  the  complementary  functions  of  its 

conn)lemont  (90°—^). 

§  24.  Tlie  lunctions  of  angles  greater  than  90°  can  be 
expressed  in  terms  of  the  functions  of  angles  less  than  90°, 
b}-  the  aid  of  the  formulas  of  §§  31  and  32,  Plane  Trigo- 
nometry. 


LOGARITHMS.  27 

Examples : 
sin  259°  25'=  sin  (180°+  79°  25')  =  -sin  79°  25'=  -  0.9<S30 
cos259°25'=cos(180°+79°25')  =  -cos79°25'=-0.1836 
sin  123°  32'=      0.8336  ctu258°19'=      0.20G8 

tan  123°  32'  =  -  1.509  sin  258°  19'  =  -  0.9792 

sec  300°  18'=      1.982  cos  174°  12' =  - 0.9949 

cos  300°  18'=      0.5045         ctn  174°  12'  =  -  9.85. 

Find  A  m  each  of  the  following  cases  :  — 

(1)  sin.l=      0.5G21  ;  (7)  esc  ^1  =      2.842; 

(2)  sin  A  =  -  0.9021  ;  (8)  esc  A  =  -  1 .4G9  ; 

(3)  tan.l=      1.451;  (9)  ctn.l=      0.4592; 

(4)  tan.l  =  -0.G921  ;  (10)  ctn  .1  =  - 0.9221  ; 

(5)  sec^l=      1.0084;  (11)  cos  yl  =      0.4G82 ; 

(6)  sec^=-2.9G4;  (12)  cos .1  =  - 0.8196. 
Answers  : 

(1)  34°  12' or  145°  48';       (7)      20°  36' or  159°  24' ; 

(2)  244°  26' or  295°  34' ;       (8)    222°  54' or  317°  06' ; 

(3)  55°  26' or  235°  26';       (9)      24°  40' or  204°  40' ; 

(4)  145°  19' or  325°  19';     (10)    132°  39' or  312°  39' ; 

(5)  7°  25' or  352°  35';     (11)      62°  05' or  297°  55' ; 

(6)  109°  43'  or  250°  17' ;     (12)    145°  02'  or  214°  58'. 

§  25.  "When  logarithms  were  invented  the}'  were  called 
artificial  numbers,  and  the  originals,  for  which  logarithms 
were  computed,  were  accordingly  called  natural  numbers." 
Dk  MouctAN. 

The  tables  just  explained,  §§  20-24,  are  therefore  called 
tables  of  the  natural  functions  of  an  angle. 

§  26.   Logarithms  of  the  Trigonometric  Functions. 

Having  found  the  natural  functions  of  an  angle  by  the 
methods  just  explained,  we   can  find  the  logarithms  of 


28  LOGARITHMS. 

these  functions  from  the  table  of  the  logarithms  of  num- 
bers (pages  2-0  of  the  Tables) :  it  more  often  happens, 
however,  that  the  natural  functions  themselves  are  of  no 
service  to  the  computer ;  he  generally  wishes  only  the 
logarithms  of  these  functions.     If,  for  example, 

a  =  16  sin  28°  esc  32°, 
we  have  log  a  =  log  1 G  +  log  sin  28°  +  log  esc  32°  ; 
and  hence,  in  order  to  fuid  the  value  of  a  we  do  not  need 
sin  28°  and  esc  32°  but  their  logarithms.  For  the  above 
reason  tables  have  l)een  constructed  (see  pages  8-15  of  the 
Tables) ,  liy  the  aid  of  which  the  logarithms  of  the  trigono- 
metric functions  can  be  found  directbj  from  the  angles. 

§  27.  Angles  between  G°  00'  and  8i°  00'  inclusive. 
(Tables:    pages  10-1.k) 

To  the  student  who  is  familiar  with  the  tables  already 
discussed  [Tables  :  pages  2-5  and  pages  22-27]  only  two 
things  need  be  explained  in  regard  to  pages  10-15. 

First.  In  order  to  avoid  the  use  of  negative  cliaracter- 
istics,  each  logarithm  whose  characteristic  is  negative  has 
been  increased  by  10.  Thus,  from  page  10  we  find 
log  sin  7°  00' =  9.0859,  and  this,  the  tabular  logarithm  of 
sin  7°  00',  is  10  too  large. 

.".  true  log  sin  7°  00'  =  tabular  log  sin  7°  00'  —  10 
=  9.0859-10  =  1.0859. 

Now,  the  characteristic  of  the  logarithm  of  any  trigono- 
metric function  will  be  negative  when  the  function  itself 
is  less  than  1,  and  positive  when  the  function  is  greater 
than  1  ;  and  we  know  either  from  Trigonometry  or  froui 
the  tables  oH  natural  functions,  pages  22-27  of  the  tables, 
that  the  sines  and  cosines  of  all  angles,  the  tangents  of 
angles  between  0°  and  15°,  and  the  cotangents  of  angles 


LOGAEITIIMS.  29 

between  45°  and  00°,  arc  less  than  1,  while  the  secants  and 
cosecants  of  all  augk's,  the  tangents  of  angles  between 
45°  and  90°,  and  the  cotangents  of  angles  between  0°  and 
45°,  are  greater  tlian  I.  Consequeutl}',  the  tabular  loga- 
rithms of  all  sines  and  cosines,  of  the  tangents  of  angles 
between  0°  and  45°,  and  of  the  cotangents  of  angles  be- 
tween 45°  and  1)0°,  are  10  too  large. 

Examx)les :  true  log  sin  81°  20'  =  9.9950  -  10 
true  logtan81°  20'  =  0.81G9 
true  log  tan  42°  10'  =  9.9570  -  10 
true  logctn42°  10'  =  0.0930. 

Second.  The  numbers  printed  in  smaller  type  are  dif- 
ferences to  be  used  in  interpolating ;  we  find,  for  example, 
between  log  sin  7°  00'  and  log  esc  7°  00',  the  number  103  in 
smaller  type,  and  this  is  the  difterence  to  be  used  in  flnd- 
nig,  l)y  the  method  of  interpolation,  the  logarithms  of  the 
sines  and  cosecants  of  angles  between  6°  55'  and  7°  05', 
i.e.,  of  angles  that  are  nearer  to  7°  00'  than  they  are  to  the 
angle  which  precedes  or  follows  it. 

Let  us  lind  log  sin  7°  02'  and  log  esc  7°  02'. 

log  sin  7°  00'  =  9.0859  log  esc  7°  00'  =  0.9141 

.2  of  103=         21  .2  of  103=         21 


.-.  log  sin  7°  02'  =  9.0880  ;      .-.  log  esc  7°  02'  =  0.9120. 

Find  log  sua  7°  08'. 

7°  08'  is  nearer  to  7°  10'  than  it  is  to  7°  00',  therefore  we 
use  for  our  difference  the  number  100  (oi)i)osite  7°  10'). 

Iogsin7°10'  =  9.09(;i 
.2  of  100  =         20 


•.  Io2sin7°08'  =  9.0941. 


30  LOGARITHMS. 

Find  logctn  7°  08'  and  log  cos  7°  07'. 

log  ctn  7°  10'  =  0.9005  log  cos  7°  10'  =  9.9966 

.2  of  102  =20  .3  of  2  =  1 


.-.  log  ctn  7°  08'  =  0.9025  ;      .-.  log  cos  7°  07'  =  9.99G7. 

Suppose  we  wish  to  find  log  sin  7°  05' ;  this  angle  hciiig 
half  way  hetween  7°  00'  and  7°  10',  we  will  take  for  onr 
difference  lOH,  which  is  half  way  between  103  (opposite 
7°  00')  and  100  (opposite  7°  10'). 

log  sin  7°  00' =  9.0859 
.5oflOU=         51 


.-.  log  sin  7°  05' =  9.0910. 

The  student  has  donT)tless  noticed  tliat  the  number  103 
(opposite  sine  and  cosecant  of  7°  00')  is  not  the  actual 
difference  between  two  successive  logarithms  of  our  table, 

for  log  sin  7°  00'  -  log  sin  6°  50'  =  104, 

and  log  sin  7°  10'  -  log  sin  7°  00'  =  102  ; 

if  Ave  refer  to  a  7-place  table,  however,  we  find 

log  sin  7°  05' =  9.0910  082 
log  sin  G°  55'  =  9.0807   189 


difference        =0.0102  893 

and  this  difference  carried  out  to  four  })laces  we  call  103. 

103,  then,  is  a  more  accurate  difference;  for  us  to  use  \\\ 
finding  the  log  sine  and  the  log  cosecant  of  angles  whicli 
fall  between  6°  55'  and  7°  05'  than  tlie  actual  difference 
between  successiA'^e  logarithms  of  our  table  would  be.  We 
see  from  the  aboA-e  hoAv  tlie  lunnbers  in  smaller  type  iiavc; 
been  obtained  ;  tliese  numbers  neA'cr  differ  much,  and  in 
most  cases  not  at  all  from  the  differences  l)etween  succes- 
sive logarithms  of  our  table. 


LOGARITHMS. 

'61 

=  9.1959         log  cos 

14°  OG' 

=  9.9867 

=  9.9400         log  cos 

59°  04' 

=  9.7110 

=  9.9855         log  cos 

70°  15' 

=  9.5288 

§  28.    Examples: 

log  sin  9°  02' 
log  sin  62°  01' 
log  sin    75°  18' 

logsinl28°06'   =9.8959  log[-cos210°26']*=  9.9357 

log[-sin200°  16']*=  9.5395  log[-cos  99°  18']*=  9.2085 

logtan   24°  02'   =9.6493  logctn    18°  04'  =0.4865 

logtan    60°  12'   =0.2421  logctn    48°  12'   =9.9514 

log[-tanl20°  04']  =  0.2374  log  [  -  ctn    99°  02']  =  9.2013 

logtan260°  18'  =0.7672  log  ctn210°  16'  =0.2339 

log  sec    16°  02'  =0.0173  log  esc    22°  01'   =0.4261 

logsec    70°  16'  =0.4715  log  esc    48°  02'  =0.1287 

log  [-seclOO°  18']  =  0.7476  log  esc  102°  58'  =0.0112 

log  [-  sec 200°  02']  =  0.0271  log  [-  ese  216° 46']  =  0.2229 

Find  A  in  each  of  the  following  cases  :  — 

(1)  log  sin  ^  =9.6021  ;      (8)  log[-tan^]=  9.9966  ; 

(2)  log  sin  yl  =  9.8588  ;      (9)        logctn^  =  0.2174  ; 

(3)  log  [-sin^]  =  9.8120;    (10)  log  [-ctnyl]=  9.7218  ; 

(4)  logcos^  =9.6821;    (11)        logsec^l  =0.1261; 

(5)  log  cos ^  =9.8266;    (12)  log  [-sec ^1]  =  0.6879  ; 

(6)  log  [-cos^]  =  9.7692;    (13)         log  esc  ^1   =0.6428; 

(7)  log  tan  .1  =0.6868;    (14)  log  [-csc.4]  =  0.6999. 

Answers : 

or  156° 25'; 
or  133°  45'; 
or  319° 33'; 

or  298° 45'; 
or  312°  08'; 


*  See  Example  2,  pp.  19,  20,  and  note,  p.  20. 


(1) 

23°  35' 

(2) 

46°  15' 

(3) 

220°  27' 

(4) 

61°  15' 

(■5) 

47°  52' 

(6) 

126°  00' 

(7) 

78°  23' 

{^) 

135°  14' 

or 

224°  46' 

('^>) 

31°  07' 

or 

211°  07' 

(10) 

117°  47' 

or 

297°  47' 

(11) 

41°  35' 

or 

318°  25' 

(12) 

101°  50' 

or 

258°  10' 

(13) 

13°  09' 

or 

166° 51' 

(14) 

191°  31' 

or 

348°  29'. 

32  LOGARITHMS, 

Angles  between  0°  and  6°  and  between  ,84°  and  90°. 
(Tables:  pages  8,  9.) 

§  29.  When  au  angle  is  near  0°  or  90°,  the  difference 
between  successive  logarithms  of  certain  of  its  functions  is 
comparatively'  large,  and  when  the  interval  in  angle  is  as 
great  as  10',  this  difference  is  so  large  that  we  cannot 
obtain  accurate  results  b}"  the  method  of  interpolation  for 
angles  which  lie  within  the  interval.  (See  note  on  page  9, 
and  also  last  paragraph  of  page  11.) 

We  lind,  therefore,  on  pages  8  and  9  of  the  tal)les,  a  sep- 
arate table  for  angles  between  0°  and  G°  and  between  84° 
and  90°,  where  the  interval  between  the  successive  angles 
of  tlie  table  is  only  1'. 

§  30.  log  sin,  log  tan,  and  log  sec  of  angles  between  0° 
and  G°. 

((()  log  sin .  Let  us  find  log  sin  1°  35'.  At  the  top  of  the 
middle  division  of  page  8  of  the  tables  we  find  1°,  1  sin, 
1  tan,  and  1  sec  ;  running  down  the  column  headed  1  sin 
until  we  reach  the  horizontal  line  which  has  35'  on  the  left, 
we  find  8.4414,  which  is  log  sin  1°  35'. 

Find  log  sin  2°  18. C  :  in  the  third  division  of  j)age  8  of 
the  tables  we  find 

log  sin  2°  19' =8.6006, 
and  on  the  right,  in  smaller  type,  the  difference  SI : 
.4  of  31  =  12; 

.-.  logsin  2°  18'.6  =  8.6054. 

Find  log  sin  3°  30'. 2.  Ans.  8.7861 . 

(b)  log  tan.  The  log  tan  of  an  angle  between  0°  and 
6°  diflfers  little  from  its  log  sin  ;  they  are,  in  general,  the 
same,   except  in    the   last   two  figures  of  the    mantissa. 


LOGARITHMS.  33 

Therefore,  in  order  to  save  space,  onl}^  the  last  two  figures 
of  the  log  tan  are  given  in  the  table. 

Find  log  tan  1°  35' :  on  the  right  of  log  sin  1°  35',  in  the 
eohinni  headed  1  tn,  we  find  16,  the  last  two  figures  of  the 
required  mantissa  ;  therefore,  remembering  that  the  char- 
acteristic and  first  two  figures  of  the  mantissa  are  the  same 
as  in  log  sin  1°  35',  we  get  log  tan  1°  35'  =  8.4416. 

Find  log  tan  2°  18'.  Ans.  8.6038. 

Find  log  tan  5°  13'.  We  find  in  the  third  division  of 
page  9  of  the  tables,  in  the  column  headed  Itn,  the  figures 
05,  and  just  to  the  left  of  these  figures  a  star ;  this  star 
indicates  that  the  characteristic  and  first  two  figures  of  the 
mantissa  are  to  l)e  taken  7iot  from  the  logarithm  to  the  left 
of  Ob,  but  from  the  following  logarithm;  thus, 

log  tan  5°  13' =  8.9605. 

In  this  case  log  tan  5°  13'  is  a  little  above  8.9600, 
and  has  therefore  8.96  for  its  first  three  figures,  while 
log  sin  5°  13',  being  a  little  below  8.9600,  has  8.95  for  its 
first  three  figures. 

Find  log  tan  3°  42'.  Ajis.  8.8107. 

Where  there  are  two  stars  in  succession  the  first  three 
figilres  are  to  be  taken  from  the  log  sin  which  follows  the 
second  star  ;  thus,  log  tan  5°  20'  =  8.9701 , 
logtan5°35'  =  8.9901. 
Find,  by  interpolating, 

logtan0°54'.4.  Ans.  8.1994. 

'logtanr02'.8.  Ans.  8.2617. 

(c)  log  sec.  The  characteristic,  and  the  first  two  figures 
of  the  mantissa,  of  log  sec  of  any  angle  between  0°  and  6° 
are  zeros,  and  are  not  given  in  the  table. 


34 


LOGARITHMS. 


Find  log  sec  1°  35' ;  in  the  secant  (sc)  column  for  1°,  and 
opposite  35 '"on  the  left,  we  find  02,  the  last  two  figures  of 
log  sec ;  prefixing  zeros  for  the  other  figures  we  get 
log  sec  r  35' =  0.0002. 
Find   (1)  log  secO°  35' ;      (3)  log  sec  2°  18'. 
(2)  log  sec  4°  29'; 
Answers:   (1)  0.0000*;   (2)  0.0013;   (3)  0.0004. 

§  31.  log  cos,  log  ctn,  and  log  esc  of  angles  betiveen  84° 
and  90°. 

We  proceed  as  in  §  30,  except  that  the  angles  and  names 
of  the  functions  are  to  be  found  at  the  bottom  of  pages  8 
and  9  of  the  tables,  while  the  minute  column  is  to  be  found 
on  the  right,  and  all  the  columns  are  to  be  read  upwards. 
Thus  : 

logcos84°lG'  =8.9996,  logctn84°16'  =9.0017, 
logcsc84°16'  =0.0022,  logcos  89°  05'.6  =  8.1993, 
logctn89°  05'.6  =  8.1994,     log  esc  89°  05'.6  =  0.0001. 


(4)  log  tan  .1  =  9.0103; 

(5)  log  tan  .4  =  8.9903. 


§  32.     To  find  an  angle   (betiveen  0°  and  C)°)  from  its 
log  sin,  log  tan,  or  log  sec. 
Given  : 

(1)  log  sin  .4  =  8.8140; 

(2)  log  sin  .4  =  8.9116; 

(3)  logtan^l  =  8.6424; 
Find  .4  in  each  case. 

Answers:   (1)3°44'.2;   (2)  4°  40'.8  ;   (3)  2°  30'.8  ; 
(4)  5°  50'.8  ;    (5)  5°  35'.2. 

*  If  an  ancle  is  near  0°  its  secant  is  near  1.  and  tliercfore  tlie 
lo2:sec  is  near  0;  the  table  gives  log  sec  0°  35'=  0.0000,  which 
means  that  log  sec  0°  35'  ditters  from  0  by  au  amount  too  small 
to  affect  the  fourth  place  of  decimals. 


LOGARITHMS. 


35 


Given:  log  sec  ^d  =  0.0016  ;  find  ^. 

Referring  to  page  8  of  the  tables,  we  find  that  0.0016  is 
the  log  sec  of  all  angles  from  4°  51'  to  4°  59'  inclusive  ;  we 
cannot,  therefore,  find  an  angle  with  definiteness  from  its 
log  sec  when  tlie  angle  is  near  0°. 

§  33.     To  find  an  angle  (between  84°  and  90°)  from  its 
log  cos,  log  ctn,  or  log  esc. 
Given  : 

(1)  log  cos  ^1  =  9.0076 

(2)  log  cos  ^  =  8.5360 

(3)  log  ctn  ^1  =  8.4190 
Find  A  in  each  case. 

Answers:  (1)  84°  09 '.5; 
(2)  88°  01'.9  ; 


(4)  log  ctn  yl  =  8.6870; 

(5)  log  esc  yl  =  0.0005; 

(6)  log  CSC  ^1  =  0.0012. 


(3)  88°  29 '.8  ; 

(4)  87°  12'.9  ; 

(5)  Any  angle*  from  87°  O8'to  87°  23',  inclnsive. 

(6)  Any  angle  *  from  85°  40'  to  8o°  49 ',  inclusive. 


§34. 


log  CSC,  log  ctn,  and  log  cos  of  angles  between  0° 
and  6°  ;  and  log  sec,  log  tan,  and  log  sin  of  angles  between 
84°  and  90°. 

We  cannot  find  these  logarithms  directly  from  the  table  ; 
we  know  from  Trigonometry,  however,  tliat 

1  .      .  1  ,1 


csc<^ 
sec^ 


sin  ^ 
1 


ctn  ci  =  • 


tan  ^  = 


tan  </) 
1 


cos  <^ 
sin  </)  = 


1 


cos  (j>  ctn  0  '       esc  cji 

From  §  16  we  Ivnow  that  the  logarithm  of  the  reciprocal  of 
a  quantity'  is  the  colog  of  the  quantit}^  minus  10,  therefore  : 
log  CSC  cji  =  colog  sin  ^—10  log  sec  (f>  =  colog  cos  </>  — 10 
log  ctn  ^  =  colog  tan  ^—10  log  tau  c^  =  colog  ctn  (/>  —  1 0 
log  cos  cf)  —  colog  sec  ^  —  10         log  sin  <^  =  colog  esc  (/>  —  1 0 

*  An  angle  near  90°  cannot  be  found  with  defluiteuess  from 
its  lo"  CSC. 


36 


LOGARITHMS. 


Let  US  find  log  esc  2°  IG' ;  from  tlie  taV)le  (p.  S)  we  get: 
tabular*  log  sin  2°  IC  =  8.5'J72  ; 
.-.  true  log  sin  2°  16'  =  2.5972  ; 
.-.  log  CSC  2°  1G'  =  11.4028-  10  =1.4028. 
What  was  done  above  is  eqni\'alent  to  subtracting  tabular 
log  sin  from  10,  for 

log  CSC  2°  16'  =  colog  sin  2°  16'  -  10 
=  (10  -  2.r>972)-10 
=  10 -(2.5972  +  10) 
=  10  -  8.5972  (tab.  log)  =  1.4028  ; 
>•.  log  esc  2°  16'  =  10  -  tab.  log  sin  2°  16' ; 
and  in  the  same  way  it  ma}'  be  shown  that  when  ^  is  an^^ 
angle  between  0°  and  6°, 

( 1 )  log  CSC  <j!)  =  10  —  tab.  log  sin  <^, 

( 2 )  log  ctn  <;()=:  1 0  —  tab.  log  tan  <^, 
and  when  0  is  between  84°  and  90° 

(;3)  log  sec  <;^  =  10  —  tal).  log  cos  <^, 

(4)  log  tan  (ji  —  10  —  tab.  log  ctn  cjy. 

If  we  also  put 

(5)  log  cos  (^  =  10  —  tab.  log  sec  ^, 

(6)  log  sin  9^  =  1 0  —  tab.  log  esc  <^, 

we  shall  get  results  which,  although  10  too  large,  will  accord 
with  other  tabular  logarithms.      (See  §  27,  first  part.) 

After  a  little  practice  the  student  can  tell  at  a  glance 
whether  a  logarithm  is  10  too  large  ;  he  will  then  be  able 
to  use  the  following  rule  :  — 

To  get  the  logarithm  of  the  reciprocal  of  a  function  sub- 
tract the  tabular  logarithm  of  the  function  from  10,  and  the 
result  ivill  be  either  the  required  logarithm  or  one  which 
differs  from  it  by  10. 

*  See  §  27,  first  part 


LOGAEITIIMS. 


37 


Examples : 
log  CSC  0°  30' =  2.0592 
logctn  0°  30' =  2.0591 
log  cos  0°  30' =  0.0000 
]ogsec«()°  18'=  1.1902 
logtaii8G°  18'=  1.1893 
lossm8(r  18' =  9.9991 


log  CSC  2°  53'  =1.2984 
logctn  2°  53'  =  1.2979 
log  cos  2°  53'  =9.9994 
log  sec  88°  02'.4=  1.4660 
logtan88°02'.4  =  1.4657 
log  sill  88°  02'. 4  =  9.9997. 


§  35.  To  find  cm  angle  {between  0°  and  6°)  from  its 
log  CSC,  logctn,  or  log  cos;  and  an  angle  (between  84°  and 
90°)  from  its  log  cos,  log  tan,  or  log  esc. 

(1)  Given:  log  esc ^=  1.2458  ;  find  ^. 
From  §  34  (1)  log  esc  ^=  10  —  tab.  log  sin  ^  ; 

.-.  tab.logsinyl  =  10-logcsc^=  10-1.2458 
=  8.7542. 

.-.  from  the  table,  .4  =  3°  15 '.3. 

(2)  Given :  log  etn  A  =  1.444  ;  find  ^. 

A71S.  (from  §  34  (2)  and  then  from  table)  ^  =  2°  03'.5. 

Find  A  in  each  of  the  following  cases  :  — 

(3)  log  cos  .1  =  9.9988;  (G)  log  cos  ^1  =  9.9976  ; 

(4)  logcsc^l=  1.7921  ;  (7)   log  esc  J.  =  1.1111  ; 

(5)  log  etn  .4  =  1.3221  ;  (8)   logctn^  =  1.1086. 

A7iswers : 

(3)  Any  angle  from  4°  11'  to  4°  20',  inclusive  ;  * 

(4)  0°55'.2;  (5)    2°  43'.9  ; 

(6)  Any  angle  between  5°  58'  and  6°  04',  inclusive  ;* 

(7)  4°26'.4;  (8)   4°  27'.2. 

*  An  angle  near  0°  can  not  be  found  with  deflniteuess  from 
its  log  COS. 


38  LOGARITHMS. 

§  36.  Explanation  of  the  formulas  by  the  aid  of  lohich 
the  small  table,  at  the  toj)  of  jxige  10  of  the  tables,  may  be 
used. 

We  know  from  Trigonometry  tliat,  in  a  circle  with  radius 
unity,  when  ^  is  small,  sin  cf)  and  tan  <fi  are  very  nearh' 
equal  to  arc^,  or,  where  <^  has  l)een  reduced  to  minutes, 
that  sill  (fy  =  arc  </>  =  ^  arc  1 ',  approximately', 

and  tan  ^  =  arc  4>  —  4>  ^i"<^  1 ' ^  approximately  ; 

.'.  (1)  log  sin  ^  =  log<^  +  log  arc  1'.  approximately. 
(2)  logtan(/)  =  log^  +  log  arc  1',  approximately. 

We  know  also  that  sin  ^  <  arc  ^  and  that  tan^  >  arc^  ; 
therefore,  log  sin  ^  obtained  by  (1)  will  be  a  little  too 
small,  and  log  tan  (^  obtained  by  (2)  will  be  a  little  too 
large ;  let  us  denote  the  amount  b}'  which  log  sin  ^  is  too 
small  by  e  and  the  amount  b}-  which  log  tan  <^  is  too  large 
by  e',  then  we  get 

log  sin  (^  =  log  cf)  +  log  arc  1 '—  e, 
log  tan  (f)  =  log  (f)  +  log  arc  1 '  +  «'• 

Now,  denoting  logarcl'— e  by  S  and  log  arc  1'+ c' by 
T,  we  get 

I.    log  sin  (j!)  =  log  ^  +  S . 
II.    log  tan  (j>  =  log  c^  +  T. 

These  are  the  formulas  by  the  aid  of  which  the  small 
table  is  to  be  used. 

Let  us  compute  S  and  T  wlien  <^  =  0°  44'. 
From  T. ,  S  =  log  sin  ^  —  log  cf). 

and  from  II . ,  T  =  log  tan  ^  —  log  <^  ; 

logsinO° 44' =  8.1072         logtan  0°44'  =  8.1072 
log  44  =1.6435         log  44  =1.6435 

S  =  6.4637  T  =  6.4637. 


LOGARITHMS. 


39 


Let  us  now  compute  log  arc  1'. 
Arc  360°  =  2  7r>-  =  2  TT  :    .  • .  arc  1 ' 


ISO 


arcl' 


log  3.1416 
lo2  10800 


60  X  180 
0.4971 
4.0334 


3.1416 
10800  ■ 


log  arc  1'  =6.4637 
and  this  is  the  same  as  the  vakies  of  S  and  T  found  above. 
Hence  we  see  that  when  ^  =  0°  44'  (or  less) ,  S  and  T 
differ  from  log  arc  1'  hy  quantities  too  small  to  affect  the 
fourth  place  of  decimals.  In  like  manner  have  been  com- 
puted the  different  values  of  8  and  T  given  in  the  small 
table.  In  this  table  the  first  three  figures  of  S  and  T  are 
printed  in  full-faced  type  at  the  top  of  the  column,  while 
the  last  two  figui'es  are  to  be  found  l)elow.  Referring  to 
this  table  we  see  that,  as  a  small  angle  ^  increases,  S  and 
T  vary  very  slowly  and  differ  little  from  log  arc  1'  [6.4637]. 

§  37.  The  logarithms  of  the  fimcfio)ts  of  angles  near  0° 
or  90°  by  the  aid  of  the  small  table  at  the  top  of  page  10  of 
the  tables. 

I.    Angles  between  0°  and  6°. 

log  sec  (f>  is  readily  found  from  the  third  division  of  the 
small  table,  where  the  first  three  figures  of  the  logarithm 
are  printed  in  full-faced  type  at  the  top  of  the  column 
and  the  last  two  figures  are  given  below.  Thus,  to  get 
log  sec  0°  18'  we  refer  to  the  column  headed  cf)  in  the  third 
division  of  the  table,  and  find  that  0°  18',  falling  between 
0°00'  and  0°  52',  has  a  log  sec  equal  to  0.0000.  In  like 
manner  we  find 

log  sec  1°  02' =  0.0001         log  sec  4°  12' =  0.001 2 
los  sec  0°  50'  =  0.0023         los  sec  2°  50'  =  0.0005. 


40  LOGARITHMS. 

log  cos  (ji  cannot  be  found  directh'  from  the  table  ;   we 

know,  however,  that  cos  (f>  =  •>  therefore  by  §  34  we 

sec  (/) 

can  get  log  cos  0  by  first  finding  log  sec  (/>  and  tlien  sub- 
tracting the  latter  from  10.     Thus, 

log  cos  0°  18'  =  10.0000         log  cos  4°  12'  =  9.9988 
log  cos  1°  02'  =    9.9999         log  cos  5°  50'  =  9.9977. 


log  sin  <^  can  be  found  b}'  the  aid  of  the  left-hand  division 
of  the  table  and  the  formula 

log  sin  (f>  =  log  ^  +  S. 

Thus,  to  get  log  sin  0°  00'.;"),  we  find  from  the  table  of 
the  logarithms  of  numbers  log  0.,')  =  1.0990  ;  we  find  from 
the  small  table  that  for  all  angles  between  0°  00'  and  0°  51' 
S  =  6.4G37  ;  .      Y  ^.  y^ 

lo£rsin0°00'.5  =  j   ^ '•,;'': 


=  6.1627 


and,  in  like  manner, 

log  sin  2°  18'  =  log  sin  138'  = 


2.1399 
+  6.4636 


8.6035 


log  sin  1°52'.6  =  8.5151 
log  sin  0°58'.8  =  8.2331. 

log  CSC  (j>  is  not  given  in  the  small  table  ;  it  must  there- 
fore be  obtained  bv  first  finding  log  sin  </>  and  then  sub- 
tracting the  latter  from  10.      (See  §  34.) 

Examples : 

log  CSC  0°  15'  =  2.3602         log  esc  2°  18'=  1.3965. 


I 


I 


LOGARITHMS.  41 

Jog  tan  (ft  can  be  found  by  tlie  aid  of  the  second  division 
of  the  small  table  and  the  formula 

log  tan  (ji  =  log  </)  +  T. 

.ogta„2M8'  =  {^^ 

=  8.G039 
logtanl°52'.G  =  8.5154 
log  tan  0°  58'. <S  =  8.2332. 
log  ctn  <f>  is   not  given  in  the  small  table  ;   it  must  be 
ol)taiuod,  therefore,  by  first  finding  log  tan  ^  and  then  sub- 
tracting the  latter  from  10.      (See  §  3-i.) 
Examples  : 

log  ctn  0°  15'=  2.3602         log  ctn  2°  08'=  1.4289. 

II.    To  find   (ji  ivhen  the  logarithm  of  any  one  of  its 
functions  is  given. 

Examples.    Find  ^  in  each  of  the  following  cases  : 

(1)   log  sin  (^  =  8.9824. 
Using  the  formula  log  sin  <^  =  log  ^  -}-  S,  we  get 
log  ^  =  log  sin  (/)  —  S. 

Now,  from  the  small  table  we  see  that  when  log  sin  <f>  is 
between  8.9845  and  8.9498,  S  =  6.4031, 
,       ,       f      8.9824 

.  • .  log  <h  —  J. 


=  2.5193 

.-.  <^  =  330'.6=5°30'.6. 

(2)  log  sin  <^=  7.2982;  c^  =  0°  0G'.8. 

(3)  logtan<^  =  7.()947  ;    in  this  case 
7.6947 


log<^  =  logtan,^-T       ,  _  (.  ^gg- 


1.2310;     .-.  ^  =  0°17'.02. 


42  LOGARITHMS. 

(4)  logtiui</)  =  8.1624;         <^  =  0°49'.95. 

(5)  logsec</>  =  0.0012; 

(}>  may  be  an}'  angle  between  4°  10'  and  4°  20',  inclusive.* 
When  logcscc/),  logetn<^,  and  log  cos  ^  are  given,  we 
must  sul)traet  them  from  10  in  order  to  get  log  sin  ^, 
log  tan  c/),  and  logcos</),  respectively,  and  then  from  these 
latter  logarithms  <^  can  be  found,  as  in  the  preceding- 
examples. 

(6)  log  CSC  <^  =  2.2352;  <^==0°20'. 

(7)  log ctnc^  =1.2250;  </)  =  3°24'.3. 

(8)  log  cos  </)  =  9.9997; 

(f>  may  be  any  angle  between  1°  57'  and  2°  17',  inclusive.* 

III.    Angles  between  84°  and  90°. 

When  4,  is  between  84°  and  90°  we  must  subtract  it  from 
90°,  and  then  get  by  the  methods  just  explained  the  com- 
plementary functions  of  the  remainder. 

Thus  :  log  sin  8G°  18'  =  log  cos  3°  42'  =  9.9991 
log  tan  86°  18' =  logctn3°  42' =  1.1893 
log  see  86°  18'  =  log  esc  3°  42'  =1.1902 
log  ctu  86°  18'  =  log  tan 3°  42'  =  8.8107 
log  CSC  86°  18'  =  log  sec  3°  42'  =  0.0009 
log  cos  86°  18'  =  log  sin  3°  42'  =  8.8098. 

§  38.  TJie  advantage  to  be  <jained  by  using  the  small 
table  at  the  top  of  page  10  0/  the  tables  for  angles  near  0° 
or  90°. 

AYlicn  an  angle  is  given  only  to  the  nearest  minute, 
nothing  can  be  gained  by  using  the  small  table,  for  in  such 
cases  the  logarithms  can  be  taken  directly  from  pages  8 
and  9.     Wlien.  however,  an  angle  which  differs  from  0°  or 

*  An  aiiirle  near  0°  cannot  be  found  with  defiuiteness  from  its 
log  sec,  or  from  its  log  cos. 


LOGARITHMS. 


43 


90°  b3^1ess  than  1°  is  given  to  decimals  of  a  minute,  greater 
accuracy  can  be  obtained  b}'  using  tlie  small  table.  Let  us 
take  the  angle  0°  04'. 8  for  example.  From  the  small  ta])le 
log  sin  0°  04'. 8  =  7.1433,  and  this  result  is  accurate  to  the 
fourth  place  of  decimals  ;  whereas  from  page  8,  by  the 
method  of  interpolation,  log  sin  0°  04'. 8  =  7.1449. 

If  we  take  an  angle  which  differs  from  0°  or  90°  b}'  less 
than  1',  0°  00'. G  for  example,  we  are  compelled  either  to  use 
the  small  table  or  to  express  our  angle  only  to  the  nearest 
minute.  By  the  small  table  we  get  log  sin  0°  00'. fi  =  G.2419, 
whereas,  using  0°  01'  for  0°  00 '.G,  we  get  from  page  8 
log  sin  0°  01'  =  G.4G37,  thereby  making  an  error  of  0.2218. 
Angles  near  0°  or  90°  are  of  frequent  occurrence,  especially 
in  Astronomical  work,  and  it  is  often  necessaiy,  even  when 
using  4-place  tables,  to  express  such  angles  to  decimals  of 
a  minute. 


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